Question

The standard reduction potential of $Pb$ and $Zn$ electrodes are $-0.126$ and $-0.763$ volts respectively. The e.m.f. of the cell $Zn \left| Zn ^{2+}(0.1 M ) \| Pb ^{2+}(1 M )\right| Pb$ is

• A. 0.637 V
• B. < 0.637 V
• C. > 0.637 V
• D. 0.889

Answer 1

Answer: (C)

$E=E^{\circ}-\frac{0.059}{n} \log \frac{\left[Z n^{+2}\right]}{\left[P b^{+2}\right]}$
$E^{\circ}=E_{c}^{\circ}-E_{a}^{\circ}=-0.126-(-0.763)$
$\Rightarrow -0.126+0.763=0.637$
$\Rightarrow 0.637-\frac{0.059}{2} \log \frac{0.1}{1}$
$=0.637-\frac{0.059}{2} \log \left(10^{-1}\right)$
$=0.637-\frac{0.059}{2}(-1) \log 10$
$\Rightarrow 0.637+\frac{0.059}{2} \times 1=0.637+0.0295$
$=0.6665$